Question

Explain the working of P-N junction diode. Draw the circuit of full wave rectifier and explain its working.

Hint:

For a P-N junction, remember: \textbf{F}orward bias \textbf{F}acilitates \textbf{F}low (low resistance). \textbf{R}everse bias \textbf{R}esists \textbf{R}un (high resistance). For a full-wave rectifier, the key is that no matter which half of the AC cycle is active, the current is always directed through the load resistor in the same direction.

Answer 1

Part 1: Working of a P-N Junction Diode
Step 1: Unbiased P-N Junction:
A P-N junction is formed by joining a p-type semiconductor (with majority carriers as holes) and an n-type semiconductor (with majority carriers as electrons). Due to the concentration gradient, electrons diffuse from the n-side to the p-side, and holes diffuse from the p-side to the n-side. This diffusion leaves behind immobile positive ions on the n-side and immobile negative ions on the p-side, creating a region devoid of mobile charge carriers called the depletion region. An electric field, called the potential barrier, is established across this region, which opposes further diffusion.
Step 2: Forward-Biased P-N Junction:
When the positive terminal of an external voltage source is connected to the p-side and the negative terminal to the n-side, the junction is forward-biased.
- The applied external electric field opposes the internal barrier field.
- If the applied voltage is greater than the barrier potential (approx. 0.7V for Si), the barrier is overcome.
- The width of the depletion region decreases.
- Majority charge carriers (holes from p-side and electrons from n-side) can now easily cross the junction.
- This results in a large current, called the forward current, flowing through the diode. The diode offers very low resistance in this state.
Step 3: Reverse-Biased P-N Junction:
When the negative terminal of the external voltage source is connected to the p-side and the positive terminal to the n-side, the junction is reverse-biased.
- The applied external electric field is in the same direction as the internal barrier field, thus strengthening it.
- The width of the depletion region increases.
- Majority charge carriers are pulled away from the junction and cannot cross it.
- A very small current, called the reverse saturation current or leakage current, flows due to the movement of minority charge carriers across the junction.
- The diode offers very high resistance in this state. This property of allowing current to flow in only one direction is called rectification.
Part 2: Full Wave Rectifier
Step 1: Circuit Diagram:
A full-wave rectifier uses two diodes and a center-tapped transformer to convert both halves of an AC input into a pulsating DC output.

Step 2: Working:
During the Positive Half-Cycle of AC Input:
- The upper end of the transformer secondary (A) is positive, and the lower end (B) is negative with respect to the center tap (T).
- Diode \(D_1\) is connected to A, so it becomes forward-biased and conducts.
- Diode \(D_2\) is connected to B, so it becomes reverse-biased and does not conduct.
- A current flows through diode \(D_1\) and the load resistor \(R_L\) in the direction from M to T.
During the Negative Half-Cycle of AC Input:
- The upper end (A) becomes negative, and the lower end (B) becomes positive with respect to the center tap (T).
- Diode \(D_1\) is now reverse-biased and does not conduct.
- Diode \(D_2\) is now forward-biased and conducts.
- A current flows through diode \(D_2\) and the load resistor \(R_L\), again in the same direction from M to T.
Conclusion:
In both halves of the input AC cycle, the current flows through the load resistor \(R_L\) in the same direction. This results in a unidirectional, pulsating DC voltage across the load. Since both halves of the AC wave are utilized, it is called a full-wave rectifier and is more efficient than a half-wave rectifier.